# Sum of finite Cosine Series

In the previous topic we discussed about the method of getting the sum of finite sine series. Now we will extend our learning to cosines. In this topic I will give a three simple and precise method how to deal with this kind of problem.

Formula: $\sum\limits_{i=1}^n cos(ix)=\displaystyle\frac{sin(\frac{nx}{2})cos(\frac{n+1}{2})x}{sin(\frac{x}{2})}$

Process:

1. Express the series in terms of summation expression
2. Determine the value of the following;
• n – number of terms which can be determined by using the sum of arithmetic sequence
• x – this is the multiplier
1. Use the formula provided and simplify the expression by applying the sum or difference of angle formula.

Sample Problem 1:

Express the sum of cos1° + cos2° + . . . + cos90° and give the sum as a single trigonometric expression.

Solution:

Following the process mentioned above

a. We can express the expression this way, $\sum\limits_{i=1}^{90} cos(i)=\displaystyle\frac{sin(\frac{nx}{2})cos(\frac{n+1}{2})x}{sin(\frac{x}{2})}$

b. Clearly, n=90 because there are 90 terms. Also it is obvious that x=1.

c. Using the formula: $\sum\limits_{i=1}^{90}cos(i)=\displaystyle\frac{sin(\frac{nx}{2})cos(\frac{n+1}{2})x}{sin(\frac{x}{2})}$ $\sum\limits_{i=1}^{90} cos(i)=\displaystyle\frac{sin(\frac{90(1)}{2})cos(\frac{90+1}{2})1}{sin(\frac{1}{2})}$ $\sum\limits_{i=1}^{90} cos(i)=\displaystyle\frac{sin(45)cos(\frac{91}{2})}{sin(\frac{1}{2})}$ $sin45=\frac{\sqrt{2}}{2}$  and $cos\frac{91}{2}=cos(45+\frac{1}{2})$ $cos(45+\frac{1}{2})=cos45cos\frac{1}{2}-sin45sin\frac{1}{2}$ $cos(45+\frac{1}{2}) = \frac{\sqrt{2}}{2}(cos\frac{1}{2}- sin\frac{1}{2})$

Going back to our formula, $\sum\limits_{i=1}^{90} cos(i)=\displaystyle\frac{(\frac{\sqrt{2}}{2}) \frac{\sqrt{2}}{2}(cos\frac{1}{2}- sin\frac{1}{2})}{sin(\frac{1}{2})}$ $\sum\limits_{i=1}^{90} cos(i)=\displaystyle\frac{cot\frac{1}{2}-1}{2}$

Sample Problem 2:

cos2+cos4+ cos6 + . . . +cos60 can be expressed as $\frac{\sqrt{A}cotB-1}{C}$ . Where x,y,z are integers. What is the value of A+B+C?

To solve this we still use the same process

The equivalent summation expression for this is $\sum\limits_{i=1}^{30} cos(2i)$

n=30 because there are only 30 terms in the series. x=2.

By substitution: $\sum\limits_{i=1}^{30} cos(2i)= \displaystyle\frac{sin(\frac{nx}{2})cos(\frac{n+1}{2})x}{sin(\frac{x}{2})}$ $\sum\limits_{i=1}^{30} cos(2i)= \displaystyle\frac{sin(\frac{30(2)}{2})cos(\frac{30+1}{2})2}{sin(\frac{2}{2})}$ $\sum\limits_{i=1}^{30} cos(2i)= \displaystyle\frac{sin(30)cos(31)}{sin(1)}$

sin30=1/2 and cos31=cos(30+1)

cos(30+1)=cos30cos1-sin30sin1

cos(30+1)=cos30cos1-sin30sin1

cos(30+1) = $\displaystyle\frac{\sqrt{3}(cos1-sin1)}{2}$

Going back to our formula: $\sum\limits_{i=1}^{30} cos(2i)=\displaystyle\frac{sin(30)cos(31)}{sin(1)}$ $\sum\limits_{i=1}^{30} cos(2i)=\displaystyle\frac{\frac{1}{2}\frac{\sqrt{3}(cos1-sin1)}{2}}{sin(1)}$

By simplifying, $\sum\limits_{i=1}^{30} cos(2i)=\displaystyle\frac{\sqrt{3}(cot1-1)}4$

Thus, A+B+C=8

Without using your calculator,find the sum of cos4+cos8+cos12+. . .+cos180